arXiv:1605.03769v1 [math.FA] 12 May 2016
RAJEEV GUPTA AND MD. RAMIZ REZA
Abstract. We show that the complex normed linear space ℓ1(n), n > 1, has no isometric embedding intok×kcomplex matrices for anyk∈Nand discuss a class of infinite dimensional operator space structures on it.
1. Introduction
In this paper, all the normed linear spaces considered are over the field of complex numbers unless specified. It is well known that there are isometric embeddings of real ℓ1(n) into real ℓ∞(k) for some k and hence into the space of k×k real matrices Mk(R). However, we prove that ℓ1(n), n > 1, has no isometric embedding into Mk for any k ∈ N.This shows that there is no operator space structure on ℓ1(n), n > 1, which can be induced by any k×k matrices A1, . . . , An. Furthermore, we study the operators space structures on ℓ1(n). We recall some definitions first.
Definition 1.1. An abstract operator space is a normed linear spaceV together with a sequence of normsk · kk defined on the linear space
Mk(V) :=
vij
|vij ∈V,1≤i, j≤k , ∀k∈N,
with the understanding that k · k1 is the norm ofV and the family of norms k · kk satisfies the compatibility conditions:
1. kT⊕Skp+q= max
kTkp,kSkq and 2. kASBkq ≤ kAkopkSkpkBkop
for all S∈Mq(V), T ∈Mp(V), A∈Mq×p(C) and B∈Mp×q(C).
Let (V,k·kk) and (W,k·kk) be two operator spaces. A linear bijectionT :V →W is said to be a complete isometry ifT⊗Ik: (Mk(V),k · kk)→(Mk(W),k · kk) is an isometry for everyk∈N. Operator spaces (V,k · kk) and (W,k · kk) are said to be completely isometric if there is a linear complete isometry T : V → W. A well known theorem of Ruan says that any operator space (V,k · kk) can be embedded, completely isometrically, into C∗-algebra B(H) for some Hilbert space H. There are two natural operator space structures on any normed linear spaceV,which may coincide. These are the MIN and the MAX operator space structures defined below.
Definition 1.2 (MIN). The MIN operator space structure denoted by MIN(V) on a normed linear space V is obtained by the isometric embedding ofV into theC∗-algebra C((V∗)1), the space of continuous functions on the unit ball (V∗)1 of the dual space V∗. Thus for vij
in Mk(V), we set
vij
M IN = sup
f(vij)
:f ∈(V∗)1 , where the norm of a scalar matrix f(vij)
is the operator norm in Mk.
The results in this paper are from the first author’s thesis “The Carath´eodory-Fej´er Interpolation Problems and the von-Neumann Inequality” submitted to the Indian Institute of Science, Bangalore-560012.
Key words and phrases. Operator Spaces, Finite dimensional embedding, MIN structure.
The first named author was supported by the NBHM, Government of India.
The second named author was supported by CSIR, Government of India.
1
Definition 1.3 (Max). Let V be a normed linear space and vij
∈Mk(V). Define
vij
M AX = sup
T vij
:T :V → B(H) ,
where the supremum is taken over all isometry mapsT and all Hilbert spacesH. This operator space structure is denoted by MAX(V).
These two operator space structures are extremal in the sense that for any normed linear space V, MIN(V) and MAX(V) are the smallest and the largest operator space structures on V respectively. For any normed linear spaceV, Paulsen [Pau92] associates a constant, namely, α(V),which is defined as following.
α(V) := sup
kIV ⊗Ikk(Mk(V),k·kMIN)→(Mk(V),k·kMAX) :k∈N .
The constant α(V) is equal to 1 if and only if V has only one operator space structure on it.
There are only a few examples of normed linear spaces for whichα(V) is known to be 1.These include α(ℓ∞(2)) = α(ℓ1(2)) = 1. In fact, it is known (cf. [Pis03, Page 77]) that α(V) > 1 if dim(V)≥3.
The map φ : ℓ∞(n) → B(Cn) defined by φ(z1, . . . , zn) = diag(z1, . . . , zn), is an isometric embedding of the normed linear space ℓ∞(n) into the finite dimensional C∗−algebra B(Cn).
Clearly, this is the MIN structure of the normed linear spaceℓ∞(n).We shall, however prove that there is no such finite dimensional isometric embedding for the dual space ℓ1(n).Nevertheless, we shall construct, explicitly, a class of isometric infinite dimensional embeddings of ℓ1(n).
Unfortunately, all of these embeddings are completely isometric to the MIN structure. In the end of this paper, using these embeddings and Parrott’s example in [Mis94], we construct an operator space structure on ℓ1(3),which is distinct from the MIN structure.
2. ℓ1(n) has no isometric embedding into any Mk
In this section, we will show that there does not exist an isometric embedding ofℓ1(n), n >1, into any finite dimensional matrix algebra Mk, k ∈N.Without loss of generality, we prove this for the case of n = 2. For the proof of the main theorem of this section, we shall need the following lemma.
Lemma 2.1. For k∈N and θ1, . . . , θk∈[0,2π), there existsa1, a2 ∈C such that
j=1,...,kmax
a1+eiθja2 <
a1 +
a2 . Proof. For any two non-zero complex numbersa1, a2,we have
j=1,...,kmax
a1+eiθja2
= max
j=1,...,k
|a1|+ei(θj+φ2−φ1)|a2| ,
whereφ1 andφ2 are the arguments ofa1 anda2 respectively. Settingαj =θj+φ2−φ1,we have
j=1,...,kmax
a1+eiθja2
2 = max
j=1,...,k
|a1|+eiαj|a2|
2
= max
j=1,...,k
|a1|2+|a2|2+ 2|a1a2|cosαj
.
Therefore
j=1,...,kmax
a1+eiθja2 =
a1 +
a2
if and only if cosαj = 1 for some j, that is, if and only ifαj = 0 for some j. Choosea1 and a2 such that φ1−φ2 6=θj for all j = 1, . . . , k. The existence of such a paira1 and a2 proves the
lemma.
The referee points out that the lemma is equivalent to the statement “There is no isometric embedding of ℓ1(n), n > 1, into ℓ∞(k) for any k ∈ N.” The argument below validates this equivalence.
SupposeS :ℓ1(2)→ℓ∞(k) defined byS(z1, z2) := (a1z1+b1z2, . . . , akz1+bkz2) is an isometry with smallest possible k ∈N. Then, due to the minimality of k,it follows that |aj|=|bj| = 1 for j = 1, . . . , k. Without loss of generality, we can assume thataj = 1 for j = 1, . . . , n. Using Lemma 2.2, we conclude that S can not be an isometry. For the converse part, we note that Lemma 2.2 is equivalent to the statement that the linear map S : ℓ1(2) → ℓ∞(k) defined by S(z1, z2) := (z1+eiθ1z2, . . . , z1+eiθkz2) can not be an isometry.
Now, we prove the main theorem of this section.
Theorem 2.2. There is no isometric embedding ofℓ1(2) into Mk for any k∈N..
Proof. Suppose there is ak−dimensional isometric embedding φof ℓ1(2). Then φis induced by a pair of operators T1, T2 ∈Mk of norm 1,defined by the rule, φ(a1, a2) =a1T1+a2T2.Let U1 and U2 inM2k be the pair of unitary maps:
Ui :=
Ti DT∗
i
DTi −Ti∗,
i= 1,2,
whereDTi is the positive square root of the (positive) operator I−Ti∗Ti.Now, we have PCk(a1U1+a2U2)|Ck =a1T1+a2T2.
(This dilating pair of unitary maps is not necessarily commuting nor is it a power dilation!) Thus ψ:ℓ1(2)→ M2k(C) defined by ψ(a1, a2) =a1U1+a2U2 is also an isometry. Since norms are preserved under unitary operations, without loss of generality, we assumeU1 =I andU2 to be a diagonal unitary, say, D. Let D = diag eiθ1, . . . , eiθ2k
. Applying Lemma 2.1, we obtain complex numbersa1 and a2 such that
j=1,...,2kmax
a1+eiθja2 <
a1 +
a2 .
Hence ψcannot be an isometry, which contradicts the hypothesis thatφ is an isometry.
Remark 2.3. Let X be a finite dimensional normed linear space. Suppose X is embedded isometrically in Mk for some k ∈ N, then the standard dual operator space structure on X∗ need not admit an embedding inMn for any n∈N.
Remark 2.4. An amusing corollary to this theorem is that the two spacesℓ∞(n) andℓ1(n) cannot be isometrically isomorphic forn >1.
Remark 2.5. Prof. G. Pisier points out that Theorem 2.2 may be true if one replaces Mk by K(H), the set of all compact operators on an infinite dimensional separable Hilbert spaceH.
3. Infinite dimensional embeddings of ℓ1(n)
In this section we construct operator space structure onℓ1(n) (n≥3) which is not completely isometric to MIN structure of ℓ1(n).
Let Hi be a Hilbert space and Ti be a contraction on Hi for i = 1, . . . , n. Assume that the unit circleT is contained inσ(Ti), the spectrum ofTi,fori= 1, . . . , n. Denote
T˜1 =T1⊗I⊗(n−1),T˜2 =I⊗T2⊗I⊗(n−2), . . . ,T˜n =I⊗(n−1)⊗Tn
and T = ( ˜T1, . . . ,T˜n).
Theorem 3.1. Suppose the operators T˜1, . . . ,T˜n are defined as above. Then, the function fT :ℓ1(n)→ B(H1⊗ · · · ⊗Hn)
defined byfT(a1, . . . , an) :=a1T˜1+· · ·+anT˜n is an isometry.
Proof. Since T ⊂σ(Ti) and Ti is a contraction for i= 1, . . . , n, it follows that T⊂ ∂σ(Ti) for i= 1, . . . , n.From (cf. [Con90, Proposition 6.7, Page 210]), we haveT⊂σa(Ti) fori= 1, . . . , n, where σa(Ti) is the approximate point spectrum of Ti.Thus for any i∈ {1, . . . , n} and λ∈T, there exists a sequence of unit vectors (xim)m∈N inHi such that
k(Ti−λ)(xim)k −→0 as m−→ ∞.
Now, applying the Cauchy-Schwarz’s inequality, we have
|h(Ti−λ)(xim),(xim)i| ≤ k(Ti−λ)(xim)kk(xim)k
=k(Ti−λ)(xim)k −→0
as m−→ ∞.Hence hTi(xim),(xim)i −→ λ asm −→ ∞. Let (a1, . . . , an) be any vector in ℓ1(n) such that none of its co-ordinates is zero. Letλ1 =e−iarg(a1), λ2 =e−iarg(a2), . . . , λn=e−iarg(an). Now for each i∈ {1, . . . , n}, we have (xim)m∈N, a sequence of unit vectors from Hi, such that
hTi(xim),(xim)i −→λi asm−→ ∞.
Asm goes to ∞, we have
h(a1T1⊗I⊗(n−1)+· · ·+anI⊗(n−1)⊗Tn)(x1m⊗ · · · ⊗xnm),(x1m⊗ · · · ⊗xnm)i
=
a1hT1(x1m),(x1m))i+· · ·+anhTn(xnm),(xnm))i −→
a1λ1+· · ·+anλn
=|a1|+· · ·+|an|=k(a1, . . . , an)k1. Hence ka1T˜1+· · ·+anT˜nk ≥ k(a1, . . . , an)k1.Also
ka1T˜1+· · ·+anT˜nk ≤ |a1|kT1k+· · ·+|an|kTnk.
Hence ka1T˜1+· · ·+anT˜nk=k(a1, . . . , an)k1,proving thatfT is an isometry.
If some of the co-ordinates in the vector (a1, . . . , an) are zero, the same argument, as above,
remains valid after dropping those co-ordinates.
An adaptation of the technique involved in the proof of Theorem3.1, also proves the following theorem.
Theorem 3.2. For i= 1, . . . , n, let Ti be a contraction on a Hilbert space Hi and T⊆ σ(Ti).
Denote T˜i=T1⊗ · · · ⊗Ti⊗IHi+1⊗···⊗Hn and T = ( ˜T1, . . . ,T˜n). Then, the function gT :ℓ1(n)→ B(H1⊗ · · · ⊗Hn)
defined by gT(a1, . . . , an) :=a1T˜1+· · ·+anT˜n is an isometry.
Remark 3.3. We show that all the operator spaces induced by the isometries defined in Theorem 3.1are completely isometric to the MIN structure.
SupposeT1, . . . , Tn are contractions on Hilbert spaces H1, . . . ,Hnrespectively with the prop- erty that T ⊆σ(Ti) for i = 1, . . . , n. Denote ˜T1 =T1⊗IH2 ⊗ · · · ⊗IHn, . . . ,T˜n = IH1 ⊗ · · · ⊗ IHn−1⊗Tn.Then the mapfT defined as in the Theorem3.1is an isometry. The dilation theorem due to Sz.-Nagy (cf. [Pau02, Theorem 1.1, Page 7]), gives unitary maps Uj : Kj → Kj, dilat- ing the contraction Tj, forj = 1, . . . , n. The operator space structure defined by the isometry g:ℓ1(n)→ B(K1⊗ · · ·⊗Kn),whereg(a1, . . . , an) =a1U1⊗IK2⊗···⊗Kn+· · ·+anIK1⊗···⊗Kn−1⊗Un, is no lesser than that of fT. Since U1, . . . , Un are unitary maps, C∗−algebra generated by U1⊗IK2⊗···⊗Kn, . . . , IK1⊗···⊗Kn−1⊗Un is commutative. From (cf. [Pis03, Proposition 1.10, Page 24]), we conclude thatg is a complete isometry.
It can similarly be shown that all the operator spaces induced by the isometries defined in Theorem3.2 are completely isometric to the MIN structure.
3.1. Operator space structures on ℓ1(n) different from the MIN structure. Parrott [Par70] provides an example of a contractive homomorphism on A(D3) which is not completely contractive. (Here A(D3) is the closure, with respect to the supremum norm on D3, of the polynomial in 3 complex variables.) Using a triple (I, U, V)(defined below) of 2×2 unitaries, it was shown in [Mis94] that examples due to Parrott may be easily thought of as examples of linear contractive maps on ℓ1(3) which are not completely contractive. Indeed this realization shows that the operator space structure on ℓ1(3) can not be unique. In this section, using the example from [Mis94], we give an explicit operator space structurek · kos onℓ1(3), which is not completely isometric to the MIN structure
Consider the following 2×2 unitary operators:
I2 =
1 0 0 1
, U :=
1 2
√3
√ 2 3 2 −12
!
andV :=
1
2 −√23
√3
2 1
2
! .
It is clear that the map h :ℓ1(3)→ M2, defined byh(z1, z2, z3) = z1I +z2U +z3V, is of norm at most 1.The computations done in [Mis94] includes the following:
(3.1) kI ⊗I+U⊗U +V ⊗Vk= 3
and
(3.2) sup
z1,z2,z3∈D
kz1I+z2U +z3Vk<3.
Choose a diagonal operator Don ℓ2(Z) such that kDk ≤1 andT⊂σ(D). Define T˜1 :=
I 0
0 D
,T˜2:=
U 0
0 D
,T˜3 :=
V 0
0 D
and ˆT1= ˜T1⊗I⊗I,Tˆ2 =I⊗T˜2⊗I, Tˆ3 =I⊗I⊗T˜n.LetS1:= ˆT1⊕I, S2:= ˆT2⊕U, S3 := ˆT3⊕V be operators on a Hilbert space K. Define S:ℓ1(3)−→B(K) by S(e1) =S1, S(e2) =S2, S(e3) = S3 and extend it linearly.
From Theorem3.1, we know that the function (z1, z2, z3)7→z1Tˆ1+z2Tˆ2+z3Tˆ3 is an isometry and since h is of norm at most 1, it follows that the map (z1, z2, z3) 7→ z1S1+z2S2+z3S3 is also an isometry. Consequently, there is an operator space structure os onℓ1(3) for which S is a complete isometry. Also from (3.1), we have
kS1⊗I +S2⊗U +S3⊗Vk ≥ kI ⊗I+U ⊗U +V ⊗Vk= 3.
Thusk(I, U, V)kos= 3,as norm of (I, U, V) is at most 3 under any operator space structure on ℓ1(3).On the other hand, from (3.2), we have
k(I, U, V)kMIN= sup
z1,z2,z3∈D
kz1I+z2U +z3Vk<3.
It follows from [Pau02, Theorem 14.1] that if there is a map φ: (ℓ1(3),MIN)→ (ℓ1(3),k · kos) which is a complete isometry, then the identity I : (ℓ1(3),MIN)→ (ℓ1(3),k · kos) must be also a complete isometry. Therefore the operator space structure k · kos is different from the MIN structure. However, althoughk(I, U, V)kMAX= 3,we are unable to decide whether the operator space structurek · kos is completely isometric to the MAX operator space structure or not.
Acknowledgement: We are very grateful to G. Misra for several fruitful discussions and suggestions. We are also thankful to the referee for many valuable suggestions.
References
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Rajeev Gupta: Department of Mathematics, Indian Institute of Science, Bangalore E-mail address: [email protected]
Md. Ramiz Reza: Department of Mathematics, Indian Institute of Science, Bangalore E-mail address: [email protected]
